Question

5. Ch4-033 LO#64 61 0.450 kg of octane burned during combustion produces 1.25 kg of carbon dioxide. What is the percent yield? 0. a. 90.1 % b. 36.0 % c. 0.901 %

Answer 1

Ans: ‘a’

Octane = C₈H₁₈

Mass of octane taken = 0.450 kg = 450 g

Molar mass of octane = 114.23 g/mol

No. of moles of octane taken = 450/114.23 = 3.939 mol

Combustion reaction: 2 C₈H₁₈ + 25 O₂ ---> 16 CO₂ + 18 H₂O

2 moles of octane give 16 moles of CO₂. 3.939 moles of octane is expected to give 31.512 moles of CO₂.

Molar mass of CO₂ = 44.01 g/mol

Mass of 31.512 moles of CO₂ = 31.512 x 44.01 = 1386.8 g    (Theoretical yield)

Mass of CO₂ formed = 1.25 kg = 1250 g (experimental yield)

% yield = (experimental yield/theoretical yield)x100 = (1250/1386.8) x 100 = 90.136